http://acm.hdu.edu.cn/showproblem.php?pid=2586
题意：求最近祖先节点的权值和
思路：LCA Tarjan算法
1 #include <stdio.h> 2 #include <string.h> 3 #define maxn 40005 4 5 int m,n,x[maxn],y[maxn],z[maxn],head[maxn*2],pos,dist[maxn],f[maxn]; 6 bool vis[maxn]; 7 struct Edge{ 8 int to,val,next; 9 }edge[maxn*2]; 10 11 12 void add(int u,int v,int val) 13 { 14 edge[pos].to = v; 15 edge[pos].val = val; 16 edge[pos].next = head[u]; 17 head[u] = pos++; 18 } 19 20 int Find(int x) 21 { 22 if(f[x]!=x) 23 return f[x] =Find(f[x]); 24 return f[x]; 25 } 26 27 void Tarjan(int u) 28 { 29 vis[u] = true; 30 f[u] = u; 31 for(int i = 1;i<=n;i++) 32 { 33 if(x[i]==u&&vis[y[i]]) z[i] = Find(y[i]); 34 if(y[i]==u&&vis[x[i]]) z[i] = Find(x[i]); 35 } 36 for(int i =head[u];i!=-1;i=edge[i].next) 37 { 38 int v = edge[i].to; 39 if(!vis[v]) 40 { 41 dist[v] =dist[u]+edge[i].val; 42 Tarjan(v); 43 f[v] = u; 44 } 45 } 46 } 47 48 int main() 49 { 50 int t,a,b,c; 51 scanf("%d",&t); 52 while(t--) 53 { 54 scanf("%d%d",&m,&n); 55 memset(head,-1,sizeof(head)); 56 memset(vis,false,sizeof(vis)); 57 58 pos = 1; 59 for(int i = 1;i<m;i++) 60 { 61 scanf("%d%d%d",&a,&b,&c); 62 add(a,b,c); 63 add(b,a,c); 64 z[i] = 0; 65 } 66 z[m] = 0; 67 for(int i = 1; i<=n;i++) 68 { 69 scanf("%d%d",&x[i],&y[i]); //分别存每一次的查询点。 70 } 71 dist[1] = 0; 72 Tarjan(1); 73 for(int i = 1;i<=n;i++) 74 printf("%d ",dist[x[i]]+dist[y[i]]-2*dist[z[i]]); 75 } 76 return 0; 77 }